Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $n = \dfrac{p^2 - 20p + 100}{-9p + 90} \times \dfrac{8p - 40}{p^2 - 10p} $
First factor the quadratic. $n = \dfrac{(p - 10)(p - 10)}{-9p + 90} \times \dfrac{8p - 40}{p^2 - 10p} $ Then factor out any other terms. $n = \dfrac{(p - 10)(p - 10)}{-9(p - 10)} \times \dfrac{8(p - 5)}{p(p - 10)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (p - 10)(p - 10) \times 8(p - 5) } { -9(p - 10) \times p(p - 10) } $ $n = \dfrac{ 8(p - 10)(p - 10)(p - 5)}{ -9p(p - 10)(p - 10)} $ Notice that $(p - 10)$ appears twice in both the numerator and denominator so we can cancel them. $n = \dfrac{ 8\cancel{(p - 10)}(p - 10)(p - 5)}{ -9p(p - 10)\cancel{(p - 10)}} $ We are dividing by $p - 10$ , so $p - 10 \neq 0$ Therefore, $p \neq 10$ $n = \dfrac{ 8\cancel{(p - 10)}\cancel{(p - 10)}(p - 5)}{ -9p\cancel{(p - 10)}\cancel{(p - 10)}} $ We are dividing by $p - 10$ , so $p - 10 \neq 0$ Therefore, $p \neq 10$ $n = \dfrac{8(p - 5)}{-9p} $ $n = \dfrac{-8(p - 5)}{9p} ; \space p \neq 10 $